A community for students.

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

anonymous

  • 5 years ago

Find all critical points of f(z)=(3z^2-6)^2(1-5z)^4

  • This Question is Closed
  1. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    z = 2 and z = 1/5

  2. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Take the derivative of f(z) with respect to z and set it equal to zero. The values of z that result will be critical points; don't know if you want to distinguish between different types though.

  3. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    I understand how to find them - I was having problems with setting the derivative to zero.

  4. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Alright; give me a minute and I'll try it out. :)

  5. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    I've got f'(z)=6(z^2-2)(6z)(1-5z)^4-15(z^2-2)4(1-5z)^3 for the derivative. Look like what you got?

  6. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    When you differentiate, it should come out to this:\[f'(z) = 6*(1-5z)^4*(3z^2-6)*2z-20*(1-5z)^3*(3x^2-6)^2\] Remember to not get sidetracked by careless errors; they're very prevalent in this problem.

  7. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Sorry, accidentally stuck an x into the last factor when it was meant to be a z :P

  8. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    The image cuts off at (3x^2 - 6)...is that the end? Just making sure

  9. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    I get a pretty simple quadratic equation in the end..yep.

  10. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Hmm..How did you get to the quadratic equation? Sorry I don't understand

  11. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    I just moved the entire right term (starting with -20*(1-5z)^3...) to the left hand side of the equation, and started simplifying. Things begin cancelling immediately and it breaks down into a quadratic equation (or so I believe)

  12. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    You set them equal to each other and simplify? How can they simplify if they are being multiplied to each other - I though you can't do that.

  13. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    No...I took that derivative and set it equal to zero, then when I moved the term beginning with -20 to the opposite side (where the 0 would be) I began cancelling common factors.

  14. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Did you get 0 = -60z^2 +12z - 20 at some point?

  15. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Did you get 0 = -60z^2 +12z - 20 at some point?

  16. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Hm...let me show you what I did: \[0 = 6*(1-5z)^4*(3z^2-6)* (2z) - 20 * (1-5z)^3*(3z^2-6)^2\]\[20 * (1-5z)^3*(3z^2-6)^2= 6*(1-5z)^4*(3z^2-6)* (2z)\](cancel like terms and move coefficients)\[\frac{5}{3}*(3z^2-6) = (1-5z)*z\]\[5z^2-10=z-5z^2\]\[10z^2-z-10 = 0\] and you can solve after that.

  17. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Ohhhhh...Okay. Thank you very much! I appreciate your help.

  18. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Glad to help. :)

  19. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    The first person to answer this was incorrect, right? You should get some weird decimal numbers after using quadratic formula?

  20. Not the answer you are looking for?
    Search for more explanations.

    • Attachments:

Ask your own question

Sign Up
Find more explanations on OpenStudy
Privacy Policy

Your question is ready. Sign up for free to start getting answers.

spraguer (Moderator)
5 → View Detailed Profile

is replying to Can someone tell me what button the professor is hitting...

23

  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • You have blocked this person.
  • ✔ You're a fan Checking fan status...

Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.